高斯分布的四阶矩

高斯分布的四阶矩

零均值高斯分布随机变量 $X\sim \mathcal{N}\left(0,{\sigma }^2\right)$ 的四阶原点矩为
$$\mathbf{E}\left[X^4\right]=3{\sigma }^4。$$
证明：
$X^4$ 的期望的计算公式为
$$\mathbf{E}\left[X^4\right]={\int }_{-\infty }^{\infty }{x}^4\frac{1}{\sqrt{2\pi }\sigma }\exp \left(-\frac{x^2}{2{\sigma }^2}\right)\mathrm{d}x\text{=}\frac{1}{\sqrt{2\pi }\sigma }{\int }_{-\infty }^{\infty }{x}^4\exp \left(-\frac{x^2}{2{\sigma }^2}\right)\mathrm{d}x\text{。}$$
根据分部积分公式
$$\int u\left(x\right)v^{\prime }\left(x\right)\mathrm{d}x=u\left(x\right)v\left(x\right)-\int v\left(x\right)u^{\prime }\left(x\right)\mathrm{d}x，$$
可以得到
$$\begin{array}{rl}{\int }_{-\infty }^{\infty }{x}^4\exp \left(-\frac{x^2}{2{\sigma }^2}\right)\mathrm{d}x=& {\int }_{-\infty }^{\infty }{x}^4\left(-\frac{{\sigma }^2}{x}\right)\left(-\frac{x}{{\sigma }^2}\right)\exp \left(-\frac{x^2}{2{\sigma }^2}\right)\mathrm{d}x\\ =& -{\sigma }^2{\int }_{-\infty }^{\infty }{x}^3\mathrm{d}\left(\exp \left(-\frac{x^2}{2{\sigma }^2}\right)\right)\\ =& -{\sigma }^2{\left(x^3\exp \left(-\frac{x^2}{2{\sigma }^2}\right)\right|}_{-\infty }^{\infty }-{\int }_{-\infty }^{\infty }3x^2\exp \left(-\frac{x^2}{2{\sigma }^2}\right)\mathrm{d}x)\\ =& 3{\sigma }^2{\int }_{-\infty }^{\infty }{x}^2\exp \left(-\frac{x^2}{2{\sigma }^2}\right)\mathrm{d}x，\end{array}$$
其中最后一步，对于第一项，有
$${x^3\exp \left(-\frac{x^2}{2{\sigma }^2}\right)|}_{-\infty }^{+\infty }=\underset{x\to +\infty }{\lim }\frac{x^3}{\exp \left(-\frac{x^2}{2{\sigma }^2}\right)}-\underset{x\to -\infty }{\lim }\frac{x^3}{\exp \left(-\frac{x^2}{2{\sigma }^2}\right)}$$
再以上式第一项为例，利用洛必达法则，
$$\begin{array}{rl}\underset{x\to +\infty }{\lim }\frac{x^3}{\exp \left(\frac{x^2}{2{\sigma }^2}\right)}=& \underset{x\to +\infty }{\lim }\frac{3x^2}{\frac{2x}{2{\sigma }^2}\exp \left(\frac{x^2}{2{\sigma }^2}\right)}=\underset{x\to +\infty }{\lim }\frac{3{\sigma }^{2}x}{\exp \left(\frac{x^2}{2{\sigma }^2}\right)}\\ =& \underset{x\to +\infty }{\lim }\frac{3{\sigma }^2}{\frac{2x}{2{\sigma }^2}\exp \left(\frac{x^2}{2{\sigma }^2}\right)}=0\end{array}$$
同理可得
$${x^3\exp \left(-\frac{x^2}{2{\sigma }^2}\right)|}_{-\infty }^{+\infty }=0$$
又有
$${\int }_{-\infty }^{\infty }{x}^2\frac{1}{\sqrt{2\pi }\sigma }\exp \left(-\frac{x^2}{2{\sigma }^2}\right)\mathrm{d}x=\mathbf{E}\left[X^2\right]={\sigma }^2，$$
因此可以得到
$$\mathbf{E}\left[X^4\right]=3{\sigma }^4。$$